Let A, B, and C be three events such that P(A | vedclass

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(C) We are given the following probabilities:$P(A \cap \bar{B} \cap \bar{C}) = 0.6$$P(A) = 0.8$$P(\bar{A} \cap B \cap C) = 0.1$We know that $P(A) = P(

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$0.3$

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(C) We are given the following probabilities:$P(A \cap \bar{B} \cap \bar{C}) = 0.6$$P(A) = 0.8$$P(\bar{A} \cap B \cap C) = 0.1$We know that $P(A) = P(A \cap \bar{B} \cap \bar{C}) + P(A \cap B \cap \bar{C}) + P(A \cap \bar{B} \cap C) + P(A \cap B \cap C)$.Substituting the known values:$0.8 = 0.6 + P(A \cap B \cap \bar{C}) + P(A \cap \bar{B} \cap C) + P(A \cap B \cap C)$$P(A \cap B \cap \bar{C}) + P(A \cap \bar{B} \cap C) + P(A \cap B \cap C) = 0.8 - 0.6 = 0.2$.The probability of at least two events occurring is given by:$P(	ext{at least two}) = P(A \cap B \cap \bar{C}) + P(A \cap \bar{B} \cap C) + P(\bar{A} \cap B \cap C) + P(A \cap B \cap C)$.Substituting the sum calculated above and the given value for $P(\bar{A} \cap B \cap C)$:$P(	ext{at least two}) = 0.2 + 0.1 = 0.3$.

Let $A$, $B$, and $C$ be three events such that $P(A \cap \bar{B} \cap \bar{C}) = 0.6$, $P(A) = 0.8$, and $P(\bar{A} \cap B \cap C) = 0.1$. Then, the value of $P(\text{at least two among } A, B, \text{ and } C)$ equals:

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